How do you write the partial fraction decomposition of the rational expression $\frac{1}{x^{3} - 6 x^{2} + 9 x}$ $1/(x^3-6x^2+9x)$ ?

Question

1558 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-14T03:53:53

$\frac{1}{9 x} - \frac{1}{9 (x - 3)} + \frac{1}{3 {(x - 3)}^{2}}$ $1/(9x)-1/(9(x-3))+1/(3(x-3)^2$

Factor the denominator.

$\frac{1}{x {(x - 3)}^{2}} = \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{{(x - 3)}^{2}}$ $1/(x(x-3)^2)=A/x+B/(x-3)+C/(x-3)^2$

$1 = A {(x - 3)}^{2} + B (x^{2} - 3 x) + C (x)$ $1=A(x-3)^2+B(x^2-3x)+C(x)$

$1 = A x^{2} - 6 A x + 9 A + B x^{2} - 3 B x + C x$ $1=Ax^2-6Ax+9A+Bx^2-3Bx+Cx$

$1 = x^{2} (A + B) + x (- 6 A - 3 B + C) + 1 (9 A)$ $1=x^2(A+B)+x(-6A-3B+C)+1(9A)$

Thus, ${(A+B=0),(-6A-3B+C=0),(9A=1):}$

Solve to see that ${(A=1/9),(B=-1/9),(C=1/3):}$

Therefore,

$1/(x^3-6x^2+9x)=1/(9x)-1/(9(x-3))+1/(3(x-3)^2$

How do you write the partial fraction decomposition of the rational expression 1/(x^3-6x^2+9x) 1x3−6x2+9x 1/(x^3-6x^2+9x) ?